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a: Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right)\cdot\dfrac{\sqrt{a}-1}{a^2}\)
\(=\dfrac{4a-1}{\sqrt{a}-1}\cdot\dfrac{\sqrt{a}-1}{a^2}\)
\(=\dfrac{4a-1}{a^2}\)
b: Để P=3 thì \(4a-1=3a^2\)
\(\Leftrightarrow3a^2-4a+1=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a-1\right)=0\)
hay \(a=\dfrac{1}{9}\)
a) ĐK: a>0; a≠1
Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right).\dfrac{\sqrt{a}-1}{a^2}\)
\(=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right).\dfrac{\sqrt{a}-1}{a^2}\)
\(=\dfrac{4a-1}{\sqrt{a}-1}.\dfrac{\sqrt{a}-1}{a^2}=\dfrac{4a-1}{a^2}\)
b) Ta có: \(P=3\Leftrightarrow\dfrac{4a-1}{a^2}=3\Leftrightarrow3a^2=4a-1\Leftrightarrow3a^2-4a+1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(loại\right)\\a=\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)
a: Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}}+1\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2}{1-\sqrt{a}}\)
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
\(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{a-1}\)
\(=\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-a-\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b: \(P=\sqrt{a}+7\)
=>\(2\left(a+2\sqrt{a}+1\right)=a+7\sqrt{a}\)
=>\(2a+4\sqrt{a}+2-a-7\sqrt{a}=0\)
=>\(a-3\sqrt{a}+2=0\)
=>\(\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>\(\left[{}\begin{matrix}a=1\left(loại\right)\\a=4\left(nhận\right)\end{matrix}\right.\)
c: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}=\dfrac{2a-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}\right)}{\sqrt{a}}=\dfrac{2\left[\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}{\sqrt{a}}>0\)
=>P>6
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a\ne1\end{matrix}\right.\)
b: Sửa đề: \(C=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left[1:\dfrac{a+\sqrt{1}-\sqrt{a}}{\sqrt{a}+1}\right]\cdot\left[\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\dfrac{a-1}{a+1}\)
c: Để C là số nguyên thì \(a-1⋮a+1\)
=>\(a+1-2⋮a+1\)
=>\(-2⋮a+1\)
=>\(a+1\in\left\{1;-1;2;-2\right\}\)
=>\(a\in\left\{0;-2;1;-3\right\}\)
Kết hợp ĐKXĐ, ta được: a=0
Bài 4. Cho biểu thức: P = ((a + 2sqrt(a))/(sqrt(a) + 2) - 1) / ((a - sqrt(a))/(sqrt(a) - 1) + 1)